$\text F = \left[\begin{array}{rr}2 & 3 \\ 0 & 5 \\ -1 & -1\end{array}\right]$ and $\text D = \left[\begin{array}{rr}-1 & 0 \\ 4 & 2\end{array}\right]$ Let $\text {H = FD}$. Find $\text H$. $ {H = }$
The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{F}$ and the first column of $\text{D}$. $ \text {H}=\left[\begin{array}{rr}{2} & {3} \\ 0 & 5 \\ -1 & -1\end{array}\right]\left[\begin{array}{rr} {-1} & 0 \\ {4} & 2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(2,3)\cdot(-1,4)\\\\ &=2 \cdot -1 + 3\cdot 4\\\\ &=10 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0 \cdot -1 + 5\cdot 4 = 20$ (Choice B) B $2 \cdot 0 + 3\cdot 2 = 6$ (Choice C) C $0 \cdot -1 + 5\cdot 0 = 0$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}= \left[\begin{array}{rr}10 & 6 \\ 20 & 10 \\ -3 & -2\end{array}\right] $